The GCD of given numbers is 1.
Step 1 :
Divide $ 6515 $ by $ 756 $ and get the remainder
The remainder is positive ($ 467 > 0 $), so we will continue with division.
Step 2 :
Divide $ 756 $ by $ \color{blue}{ 467 } $ and get the remainder
The remainder is still positive ($ 289 > 0 $), so we will continue with division.
Step 3 :
Divide $ 467 $ by $ \color{blue}{ 289 } $ and get the remainder
The remainder is still positive ($ 178 > 0 $), so we will continue with division.
Step 4 :
Divide $ 289 $ by $ \color{blue}{ 178 } $ and get the remainder
The remainder is still positive ($ 111 > 0 $), so we will continue with division.
Step 5 :
Divide $ 178 $ by $ \color{blue}{ 111 } $ and get the remainder
The remainder is still positive ($ 67 > 0 $), so we will continue with division.
Step 6 :
Divide $ 111 $ by $ \color{blue}{ 67 } $ and get the remainder
The remainder is still positive ($ 44 > 0 $), so we will continue with division.
Step 7 :
Divide $ 67 $ by $ \color{blue}{ 44 } $ and get the remainder
The remainder is still positive ($ 23 > 0 $), so we will continue with division.
Step 8 :
Divide $ 44 $ by $ \color{blue}{ 23 } $ and get the remainder
The remainder is still positive ($ 21 > 0 $), so we will continue with division.
Step 9 :
Divide $ 23 $ by $ \color{blue}{ 21 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 10 :
Divide $ 21 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 11 :
Divide $ 2 $ by $ \color{blue}{ 1 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 1 }} $.
We can summarize an algorithm into a following table.
| 6515 | : | 756 | = | 8 | remainder ( 467 ) | ||||||||||||||||||||
| 756 | : | 467 | = | 1 | remainder ( 289 ) | ||||||||||||||||||||
| 467 | : | 289 | = | 1 | remainder ( 178 ) | ||||||||||||||||||||
| 289 | : | 178 | = | 1 | remainder ( 111 ) | ||||||||||||||||||||
| 178 | : | 111 | = | 1 | remainder ( 67 ) | ||||||||||||||||||||
| 111 | : | 67 | = | 1 | remainder ( 44 ) | ||||||||||||||||||||
| 67 | : | 44 | = | 1 | remainder ( 23 ) | ||||||||||||||||||||
| 44 | : | 23 | = | 1 | remainder ( 21 ) | ||||||||||||||||||||
| 23 | : | 21 | = | 1 | remainder ( 2 ) | ||||||||||||||||||||
| 21 | : | 2 | = | 10 | remainder ( 1 ) | ||||||||||||||||||||
| 2 | : | 1 | = | 2 | remainder ( 0 ) | ||||||||||||||||||||
| GCD = 1 | |||||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.